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yulyashka [42]
2 years ago
8

Which cellular process produces simple sugars?

Chemistry
2 answers:
Nataly_w [17]2 years ago
7 0

Answer:

The answer is C. Hope this helps you out!

dexar [7]2 years ago
6 0

Answer:

C. Respiration

✂---------------

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How many grams are of sodium chlorate are present in a 2.45 mole sample?
DedPeter [7]

Answer:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles NaCl, or 58.44277 grams.

Explanation:

I hope this helps!!

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2 years ago
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24 POINTS!!!!!!!!!!
Oliga [24]
Your answer is going to be A, because it was shoved harder, it will go faster

8 0
3 years ago
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Both of the sulfur-oxygen double bonds in so2 is polar. in which direction should the polarity arrows point
kodGreya [7K]

Answer:

Away from the central sulfur atom.

Explanation:

4 0
3 years ago
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True or False: In every gram of sucrose (table sugar) there is 0.513 g of oxygen
Ber [7]

Answer:

True

Explanation:

Step 1: Find molecular formula of sucrose

C₁₂H₂₂O₁₁

Step 2: Convert moles of oxygen present to grams

1 mol O = 16 g O

11 mol O = 176 g O

Step 3: Find molar mass of sucrose

C - 12.01 g/mol

H - 1.01 g/mol

O - 16.00 g/mol

12.01(12) + 22(1.01) + 11(16.00) = 342.34 g/mol C₁₂H₂₂O₁₁

Step 4: Set up dimensional analysis

1 gC_{12}H_{22}O_{11}(\frac{176 gO}{342.34gC_{12}H_{22}O_{11}} )

Step 5: Multiply/Divide and cancel out units

Grams of C₁₂H₂₂O₁₁ and grams of C₁₂H₂₂O₁₁ cancel out.

We are left with grams Oxygen

176/342.34 = 0.514109 grams Oxygen

4 0
3 years ago
A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i
Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C  

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\

\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}

\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}

The final temperature of the water would be 35.5 °C.

7 0
3 years ago
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